# How to read bra-ket notation? [closed]

Good afternoon, I am trying to understand the basics of some quantum mechanics theorems (e.g. Uncertainity principle). I'm looking for the correct way to read this expression while I'm speaking. For instance $$\langle a|b\rangle$$ or

$$\langle(A-\langle A\rangle) \psi|(A-\langle A\rangle)\psi\rangle$$

And, more in general, what is the basic difference between $$\langle a|$$ and$$|a\rangle$$?

• I think this question is just too broad: the answer is a chapter in any QM book. I particularly recommend Shankar's. – Javier Sep 2 '19 at 13:55
• Did you have a look at the Wikipedia page of bra-ket notation? Does this answer your question or can you re-formulate it to be more specific? – ahemmetter Sep 2 '19 at 13:56
• Many thanks. I looked at it but I have some doubt on how to properly read with my voice the longest expression I posted. – muserock92 Sep 2 '19 at 13:57
• Wait, do you mean how to read out loud the expressions? I'm not sure that's really a physics question; your last question about the difference between bras and kets is one, but it's probably too broad for the site. – Javier Sep 2 '19 at 14:12
• I was thinking about how to tell this expression if I have to speak with someone. – muserock92 Sep 2 '19 at 14:15

• $$\ket{A}$$ is a column vector containing the probability amplitudes of the system
• $$\bra{A}$$ is a row vector, and the complex transpose of $$\ket{A}$$, i.e. $$\bra{A} = \ket{A}^{\dagger}$$. It comes in handy when we want to calculate the expected value of a measurement outcome. See it as a mathematical tool.
• $$\left$$ is the inner product of $$A$$ and $$B$$, which gives you a scalar
• $$\ket{A}\bra{B}$$ is the outer product of $$A$$ and $$B$$, which gives you a matrix (called the density matrix)
For two-level quantum systems such as a qubit, this way of thinking is very intuitive, since $$\ket{A}$$ simply becomes a column vector with 2 entries. However, when the systems are continuous, we suddenly deal with infinite dimensional vectors but essentially the logic is the same.